A galvanometer of resistance $100 \Omega$ is connected to a battery of $2 \mathrm{~V}$ with a resistance of $1900 \Omega$ in series. The deflection obtained is 30 divisions. To reduce this deflection by 10 divisions the additional resistance required to be connected in series isA. $1500 \Omega$B. $500 \Omega$C. $1000 \Omega$D. $2000 \Omega$

Home » A galvanometer of resistance is connected to a battery of with a resistance of in series. The deflection obtained is 30 divisions. To reduce this deflection by 10 divisions the additional resistance required to be connected in series isA. B. C. D.

A galvanometer of resistance $100 \Omega$ is connected to a battery of $2 \mathrm{~V}$ with a resistance of $1900 \Omega$ in series. The deflection obtained is 30 divisions. To reduce this deflection by 10 divisions the additional resistance required to be connected in series is
A. $1500 \Omega$
B. $500 \Omega$
C. $1000 \Omega$
D. $2000 \Omega$

Physics

A galvanometer of resistance $100 \Omega$ is connected to a battery of $2 \mathrm{~V}$ with a resistance of $1900 \Omega$ in series. The deflection obtained is 30 divisions. To reduce this deflection by 10 divisions the additional resistance required to be connected in series is
A. $1500 \Omega$
B. $500 \Omega$
C. $1000 \Omega$
D. $2000 \Omega$

Correct option is C.

Total resistance in the circuit is $R=1900+100=2000 \Omega$
$\therefore$ Current $I_{1}=\frac{2}{2000}=10^{-3} A$
The deflection is decreased from 30 to 20 divisions $\therefore I_{2}=\frac{2}{3} I_{1}=\frac{2}{3} \times 10^{-3} A$
If total resistance now is $\mathrm{R}$ ‘ then
$R^{\prime}=\frac{V}{I_{2}}=\frac{2 \times 3}{2 \times 10^{-3}}=3 \times 10^{3} \Omega=3000 \Omega$
$\therefore$ Additional resistance $=R^{\prime}-R=3000-2000=1000 \Omega$

i

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